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A researcher plans to identify each participant in a certain medical

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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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New post 12 Jul 2019, 09:20
1
A,B,AB,C,AC,BC,D,AD,BD,CD,E,AE....

5 Letters and we can get 12 code in alphabetical order.

Is it the right way of doing it?
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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New post 12 Jul 2019, 10:43
Hi sk710,

Based on the work that you've shown, you used a 'brute force' approach (re: you just 'mapped out' the possibilities without the need of a formula or any excessive 'math'). That's actually a great way to approach this question - and you'll find that a certain number of Quant questions (including many DS questions) can be solved by taking that same general approach.

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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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New post 02 Apr 2020, 07:51
Bunuel wrote:
eaakbari wrote:
Bunuel wrote:

Practice: try to use the same concept.


Okay here goes,

The # of single letter codes possible would be \(n\) itself;
The # of pair of distinct letters codes possible would be (in alphabetical order); \(nC2\)
The # of Triples of distinct letters codes possible would be (in alphabetical order); \(nC3\)

Thus

\(nC3 + nC2 + n\)> \(12\)

\(n*(n-1)/2 + n*(n-1)*(n-2)/3*2 + n\)> \(12\)

Simplifying

\(n*(n^2 +5)\)> \(72\)

Only sufficient value of \(n = 4\)

Is it correct?


Correct.

Three letters A, B, and C, are enough for 7<12 codes:
A;
B;
C;
AB;
AC;
BC;
ABC.

Four letters A, B, C, and D are enough for 15>12 codes:
A;
B;
C;
D;
AB;
AC;
AD;
BC;
BD;
CD;
ABC;
ABD;
ACD;
BCD;
ABCD.


hi Bunnel,

one doubt, It is mentioned in the question :
a pair of alphabet

How does ABC,ABD etc satisfy this condition ?
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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New post 02 Apr 2020, 07:57
Harsh2111s wrote:
Bunuel wrote:
eaakbari wrote:
Okay here goes,

The # of single letter codes possible would be \(n\) itself;
The # of pair of distinct letters codes possible would be (in alphabetical order); \(nC2\)
The # of Triples of distinct letters codes possible would be (in alphabetical order); \(nC3\)

Thus

\(nC3 + nC2 + n\)> \(12\)

\(n*(n-1)/2 + n*(n-1)*(n-2)/3*2 + n\)> \(12\)

Simplifying

\(n*(n^2 +5)\)> \(72\)

Only sufficient value of \(n = 4\)

Is it correct?


Correct.

Three letters A, B, and C, are enough for 7<12 codes:
A;
B;
C;
AB;
AC;
BC;
ABC.

Four letters A, B, C, and D are enough for 15>12 codes:
A;
B;
C;
D;
AB;
AC;
AD;
BC;
BD;
CD;
ABC;
ABD;
ACD;
BCD;
ABCD.


hi Bunnel,

one doubt, It is mentioned in the question :
a pair of alphabet

How does ABC,ABD etc satisfy this condition ?


Have you read the highlighted part? Those are examples for additional questions answered in my post.
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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New post 16 Apr 2020, 06:25
Bunuel wrote:
brandon7 wrote:
Bunuel wrote:
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8


Say there are minimum of \(n\) letters needed, then;

The # of single letter codes possible would be \(n\) itself;
The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order);

We want \(C^2_n+n\geq{12}\) --> \(\frac{n(n-1)}{2}+n\geq{12}\) --> \(n(n-1)+2n\geq{24}\) --> \(n(n+1)\geq{24}\) --> \(n_{min}=5\).

Answer: B.

Hope it's clear.


Can someone explain what this \(C^2_n+n\geq{12}\) means? I also saw a reference to the same type of symbol with an A instead, what does that mean?


C stands for combinations: \(C^2_n=\frac{n!}{2!(n-2)!}\)

Can you please explain how you get from n!/2!(n-2)! to n(n-1)/2?

Thank you!
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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New post 16 Apr 2020, 06:36
1
Franchise wrote:
Bunuel wrote:
brandon7 wrote:
Bunuel wrote:
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8


Say there are minimum of \(n\) letters needed, then;

The # of single letter codes possible would be \(n\) itself;
The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order);

We want \(C^2_n+n\geq{12}\) --> \(\frac{n(n-1)}{2}+n\geq{12}\) --> \(n(n-1)+2n\geq{24}\) --> \(n(n+1)\geq{24}\) --> \(n_{min}=5\).

Answer: B.

Hope it's clear.


Can someone explain what this \(C^2_n+n\geq{12}\) means? I also saw a reference to the same type of symbol with an A instead, what does that mean?


C stands for combinations: \(C^2_n=\frac{n!}{2!(n-2)!}\)

Can you please explain how you get from n!/2!(n-2)! to n(n-1)/2?

Thank you!


Hi Franchise,

Definition of n!=n*(n-1)*(n-2)*(n-3).......1
Definition of (n-2)!=(n-2)*(n-3)*(n-4).......1

Now look at

\(C^2_n=\frac{n!}{2!(n-2)!}\)

Replace n! and (n-2)! with above definition, we will get

n(n-1)/2

Hope it helps.
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Re: A researcher plans to identify each participant in a certain medical   [#permalink] 16 Apr 2020, 06:36

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